Question

its urgent guys please​

Answer 1

Step-by-step explanation: First

$\dfrac{\cot^2\theta+\sec^2\theta}{\tan^2\theta+cosec^2\theta} =\dfrac{cosec^2\theta-1+\sec^2\theta}{\tan^2\theta+cosec^2\theta} =\dfrac{cosec^2\theta+\tan^2\theta}{\tan^2\theta+cosec^2\theta} =1$

and since

$1=\dfrac{\sin\theta\cos\theta}{\sin\theta\cos\theta}\\=(\sin\theta\cos\theta)\dfrac{1}{\sin\theta\cos\theta}\\=(\sin\theta\cos\theta)\dfrac{\sin^2\theta+\cos^2\theta}{\sin\theta\cos\theta}\\=(\sin\theta\cos\theta) \left(\dfrac{\sin\theta}{\cos\theta}+\dfrac{\cos\theta}{\sin\theta} \right)=(\sin\theta\cos\theta) \left(\tan\theta+\cot\theta \right)$

we conclude

$\dfrac{\cot^2\theta+\sec^2\theta}{\tan^2\theta+cosec^2\theta} =(\sin\theta\cos\theta)(\tan\theta+\cot\theta)$